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放缩不等式推导
1
)
a
x
>
x
+
1
(
1
<
a
≤
e
,
x
<
0
;
a
≥
e
,
x
>
0
)
;
1) a^x>x+1left(1<aleq e,x<0;ageq e,x>0
ight);
1) ax>x+1(1<a≤e,x<0;a≥e,x>0);
p
r
o
o
f
:
proof:
proof:
f
01
(
x
)
=
a
x
−
(
x
+
1
)
⇒
f
01
′
(
x
)
=
a
x
ln
a
−
1
f_{01}left(x
ight)=a^{x}-left(x+1
ight)Rightarrow f_{01}^{'}left(x
ight) = a ^{x} ln a-1
f01(x)=ax−(x+1)⇒f01′(x)=axlna−1
1
<
a
≤
e
,
x
<
0
⇒
0
<
a
x
<
1
,
0
<
ln
a
≤
1
⇒
f
01
(
x
)
>
f
01
(
0
)
=
1
−
1
=
0
⇒
a
x
>
x
+
1
(
1
<
a
≤
e
,
x
<
0
)
;
1<aleq e,x<0\Rightarrow0<a^{x}<1,0<ln aleq1\Rightarrow f_{01}left(x
ight)>f_{01}left(0
ight)=1-1=0\Rightarrow a^{x}>x+1left(1<aleq e,x<0
ight);
1<a≤e,x<0⇒0<ax<1,0<lna≤1⇒f01(x)>f01(0)=1−1=0⇒ax>x+1(1<a≤e,x<0);